Is cos^8(x)=sin^5(x)cos^3(x) ?

The answer to whether cos^8(x)=sin^5(x)cos^3(x) is False

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prove cos^8(x)=sin^5(x)cos^3(x)

prove

cos^{8} (x) = sin^{5} (x) cos^{3} (x)

False

Solution steps

cos^{8} (x) = sin^{5} (x) cos^{3} (x)

Show that the two sides are not equal

For

cos^{8} (x) = sin^{5} (x) cos^{3} (x)

plug in

x = 0

cos^{8} (0) = 1

cos^{8} (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= 1^{8}

Simplify

= 1

sin^{5} (0) cos^{3} (0) = 0

sin^{5} (0) cos^{3} (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= 0^{5} \cdot 1^{3}

Simplify

= 0

The two sides are not equal

\Rightarrow False

p ro v e 2 sec (2 x) = tan (\frac{π}{4} + x) + tan (\frac{π}{4} - x)

p ro v e sin (x) - csc (x) = - (cos (x)) (cot (x))

p ro v e cos (θ) = sin (3 θ + 62)

p ro v e sin (π \cdot x) + sin (π \cdot (10 - x)) = 0

p ro v e cot (2 θ) = \frac{1}{2} (tan (θ) - cot (θ))

Is cos^8(x)=sin^5(x)cos^3(x) ?
The answer to whether cos^8(x)=sin^5(x)cos^3(x) is False